3y^2-10=y+6

Solution for 3y^2-10=y+6 equation:

3y^2-10=y+6

We move all terms to the left:

3y^2-10-(y+6)=0

We get rid of parentheses

3y^2-y-6-10=0

We add all the numbers together, and all the variables

3y^2-1y-16=0

a = 3; b = -1; c = -16;
Δ = b2-4ac
Δ = -12-4·3·(-16)
Δ = 193
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-\sqrt{193}}{2*3}=\frac{1-\sqrt{193}}{6}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+\sqrt{193}}{2*3}=\frac{1+\sqrt{193}}{6}
$